2002 AMC 10P Problems/Problem 1

Problem

The ratio $\frac{(2^4)^8}{(4^8)^2}$ equals

$\text{(A) }\frac{1}{4} \qquad \text{(B) }\frac{1}{2} \qquad \text{(C) }1 \qquad \text{(D) }2 \qquad \text{(E) }8$

We can use basic rules of exponentiation to solve this problem.

$\frac{(2^4)^8}{(4^8)^2} =\frac{(2^4)^8}{(2^{16})^2} =\frac{2^{32}}{2^{32}} =1$

Thus, our answer is $\boxed{\textbf{(C) } 1}.$

2002 AMC 10P (Problems • Answer Key • Resources)
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